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Does there exist an increasing sequence $a_1,a_2,\dots$ of positive integers such that both of the following are fulfilled:

  • Each positive integer appears exactly once among $a_2-a_1,a_3-a_2,\dots$

  • For some $n$, each positive integer at least $n$ appears exactly once among $a_3-a_1,a_4-a_2,a_5-a_3,\dots$?

If we only require the first condition, we can take the sequence $1,2,4,7,11,\dots$, while if we only require the second condition, we can construct the sequence by induction, making sure that each difference appears once.

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  • $\begingroup$ Are you asking if it can be constructed by induction? Or are you saying it can be constructed by induction? $\endgroup$ – астон вілла олоф мэллбэрг Sep 12 '16 at 9:33
  • $\begingroup$ I'm saying that if we only require the second condition, the sequence can be constructed by induction. $\endgroup$ – pi66 Sep 12 '16 at 9:34
  • $\begingroup$ So your question is, can both the conditions be satisfied together? Because the question is not clear. $\endgroup$ – астон вілла олоф мэллбэрг Sep 12 '16 at 9:35
  • $\begingroup$ "such that both of the following are fulfilled" $\endgroup$ – pi66 Sep 12 '16 at 9:36
  • $\begingroup$ Ok great. I'll have a look. $\endgroup$ – астон вілла олоф мэллбэрг Sep 12 '16 at 9:36
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To fulfil the first condition, the differences $a_i-a_{i-1}$ must be distinct positive integers. Thus, to avoid repeating a difference, $$a_n\ge a_0+\frac{n(n+1)}2.$$ The second condition requires that all but finitely many positive integers appear among the $a_i-a_{i-2}$. But the above growth is too fast to enable that. Thus the second condition cannot be fulfilled.

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  • $\begingroup$ It is not so clear that the second condition cannot be fulfilled... $\endgroup$ – pi66 Oct 23 '17 at 12:12

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